# How many commuting nilpotent matrices are there?

To be precise, fix $n$, fix a field $k$.

What is the maximal dimension of a subspace of the vector space of all $n\times n$ matrices formed by commutative nilpotent matrices? By commutative I mean all the products of matrices in this subspace are commutative.

(I feel like this can be formulated in terms of Lie algebras, but I don't find a good one. And I think the down-to-earth formulation might make it more accessible.)

• Commuting matrices can be simultaneously trigonalized over the algebraic closure. If the matrices are nilpotent, their trigonalizations will be strictly upper triangular (because any nonzero entries on the diagonal would survive taking powers, contradicting the nilpotency). So the maximal dimension is $\frac{n\left(n-1\right)}{2}$. – darij grinberg Sep 18 '10 at 22:12
• I get this point, but upper triangular matrices don't necessarily commute with each other, right? – Yuhao Huang Sep 18 '10 at 22:25
• "Commutative" is not a property of a matrix, but of a set of matrices, so I would rather you state the property you want more precisely. – Qiaochu Yuan Sep 18 '10 at 22:28
• I think the substitution of "commuting" for "commutative" would address Qiaochu's concern. – Charles Staats Sep 18 '10 at 22:35
• Related questions: mathoverflow.net/questions/19591/…, mathoverflow.net/questions/19755/…. It came up there that you can achieve the maximal dimension for a commutative subalgebra, 1 plus the floor of n^2/4, by taking scalars plus 2-by-2 block strictly upper triangular matrices. (Apparently the proof of maximality is due to Schur.) Now you can throw out the multiples of the identity to obtain an example of commuting nilpotents having dimension the floor of n^2/4. – Jonas Meyer Sep 19 '10 at 0:28

Try the matrices of the form ${0A\choose 00}$ with A an m by n block (with |m-n|≤1 for maximal dimension).
A book "Commutative Matrices", Acad. Press, 1968 by Suprunenko and Tyshkevich (translation from Russian) is devoted, largely, to this question. My cursory look at it tells that the problem seems to be complicated - there are only partial results for small $n$ and particular classes of algebras, expressed in terms of some cumbersome-looking invariants.